Ask question

Ask question

All categories

strojnjashka [21]

3 years ago

13

A 75 kg football player is gliding forward across very smooth ice at 4.6 m/s. He throws a 0.47 kg football straight forward. A)

What is the player's speed afterward if the ball is thrown at 15 m/s relative to the ground?B) What is the player's speed afterward if the ball is thrown at 15 m/s relative to the player?

1 answer:

lord [1]3 years ago

4 0

Answer:

4.53482 m/s

4.506 m/s

Explanation:

$m_1$ = Mass of player = 75 kg

$v_1$ = Initial velocity of player = 4.6 m/s

$m_2$ = Mass of ball = 0.47 kg

$v_1$ = Initial velocity of ball = 15 m/s

The linear momentum of the system is conserved

$(m_1+m_2)v_1=m_1v+m_2v_2\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2v_2}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times 15}{75}\\\Rightarrow v=4.53482\ m/s$

The player's speed is 4.53482 m/s

In the second case the equation of momentum is

$(m_1+m_2)v_1=m_1v+m_2(v_2+v_1)\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2(v_2+v_1)}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times (15+4.6)}{75}\\\Rightarrow v=4.506\ m/s$

The player's speed is 4.506 m/s

You might be interested in

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te

Maksim231197 [3]

Answer:

Tt = 70 + 135e^-0.031t

13 minutes

Explanation:

Given that :

Initial temperature, Ti = 205°

Temperature after 2.5 minutes = 195°

Temperature of room, Ts= 70

Using the relation :

Tt = Ts + Ce^-kt

Temperature after time, t

When freshly poured, t = 0

205 = 70 + Ce^-0k

205 = 70 + C

C = 205 - 70 = 135°

T after 2.5 minutes to find proportionality constant, k

Tt = Ts + Ce^-kt

195 = 70 + 135e^-2.5k

125 = 135e^-2.5k

125 / 135 = e^-2.5k

0.9259 = e^-2.5k

Take In of both sides :

−0.076989 = - 2.5k

k = −0.076989 / - 2.5

k = 0.031

Equation becomes :

Tt = 70 + 135e^-0.031t

t when Tt = 160

160 = 70 + 135e^-0.031k

90 = 135e^-0.031t

90/135 = e^-0.031t

0.6667 = e^-0.031t

In(0.6667) = - 0.031t

−0.405465 = - 0.031t

t = 0.405465/ 0.031

t = 13.071

t = 13 minutes

8 0

2 years ago

A 69.5-kg person throws a 0.0475-kg snowball forward with a ground speed of 31.5 m/s. A second person, with a mass of 57.5 kg, c

Leno4ka [110]

Answer:

- After throwing the snow, velocity of the thrower is 2.33 m/s

- the velocity of the receiver is 0.026 m/s

Explanation:

Given the data in the question;

Using conservation of momentum,

Initial thrower has a momentum of mv; $m_{total$ v

(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s

Now, When he throws it at 31.5 m/s, these constitutes a momentum of;

(0.0475 kg )(31.5 m/s) = 1.49625 kg.m/s

hence his momentum now is: 163.4366 - 1.49625 = 161.94035 kg.m/s

To get his velocity, we say;

161.94035 = mv

{ he lost weight of the snow ball so, m = 69.5 kg )

161.94035 = 69.5 × v

v = 161.94035 / 69.5

v = 2.33 m/s

Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s

Next is the Receiver;

the receiver will gain momentum of 1.49625 kg.m/s

he has no momentum initially and after he catches the snow ball;

1.49625 kg.m/s = mv

1.49625 kg.m/s = ( 57.5 kg + 0.0475 kg ) × v

1.49625 kg.m/s = 57.5475 kg × v

v = ( 1.49625 kg.m/s ) / 57.5475 kg

v = 0.026 m/s

Therefore, the velocity of the receiver is 0.026 m/s

3 0

3 years ago

An object whose specific gravity is 0.850 is placed in water. What fraction of the object is below the surface of the water?

Fynjy0 [20]

Answer:

The fraction of the object that is below the surface of the water is ¹⁷/₂₀

Explanation:

Given;

specific gravity of the object, γ = 0.850

Specific gravity is given as;

$specific \ gravity = \frac{density \ of the \ object}{density \ of \ water}\\\\0.85= \frac{density \ of the \ object}{1000 \ kg/m^3} \\\\density \ of the \ object = 850 \ kg/m^3$

Fraction of the object's weight below the surface of water is calculated as;

$= \frac{850}{1000} \ \times\ 100\%\\\\= 85 \% \\\\= \frac{17}{20}$

Therefore, the fraction of the object that is below the surface of the water is ¹⁷/₂₀

8 0

3 years ago

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

7 0

3 years ago

A motorcycle has 100,000 J of kinetic energy and is traveling at 20 m/s. Find its mass.

tankabanditka [31]

Answer:

<h3>The answer is 500 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{2KE}{ {v}^{2} } \\$

v is the velocity

KE is the kinetic energy

From the question we have

$m = \frac{2 \times 100000}{ {20}^{2} } = \frac{200000 }{400} = \frac{2000}{4} \\$

We have the final answer as

<h3>500 kg</h3>

Hope this helps you

7 0

3 years ago